 In some graphs, the set of vertices v can be partitioned into n-sets, where every edge joins vertex in one set to a vertex in a different set, but no edge joins vertices in the same set. If n equals 2, this gives us a bipartite graph. For example, in chemistry, an ionic compound forms when a cation binds with an anion, but two cations or two anions can't bond. In machining, a nut can fit onto a bolt, but two nuts or two bolts can't be fit together, and there are other examples. So if we look at a graph, we can easily determine if it's bipartite unless we can't. The thing to remember is that how we draw the graph is not the graph. So even if a graph is bipartite, it might not be easy to tell from the drawing. So while it's easy to determine if a small graph is bipartite, it can be more difficult to decide if a larger graph is. To decide, let's consider a path through a bipartite graph. Can we say anything about such a path? To begin with, let's consider a tree. If we rearrange the vertices of a tree, it seems we can always partition the vertices into two sets where no edge joins a vertex in one set to a vertex in the other set. Let's prove it. Or maybe disprove it. Remember, it's the journey, not the destination. It seems that the vertices in a tree can be distinguished by whether they are an even or an odd distance from some starting vertex. So let's pick a vertex v0 and partition the vertices into two sets, v1 the vertices at an odd distance from v0, and v2 the vertices at an even distance from v0. Suppose you and v are two vertices at an odd distance from v0. Could there be an edge between them? Suppose there was. Remember that in a tree, the geodesics are unique. So if there is a last vertex w that the two geodesics have in common, giving us a path from w to u, a path from w to v, but if there's an edge from u to v, this produces a cycle, and trees can't have cycles. And this proves every tree is a bipartite graph. And since we have it, we might as well note that the vertex sets are determined by whether they are an even or an odd distance from some selected starting vertex v0. Now remember one of the aspects of proof is that it raises new questions, and developing as a mathematician means being in the habit of asking new questions. In this case, one question is, if we used a different starting vertex, would we get a different partition? And we could extend the idea of bipartite sets. So what about trees and tripartite sets? We'll leave those questions for homework. In the meantime, because trees don't have cycles and have unique geodesics, analyzing them is much easier. What if our graph has cycles? And here's a useful idea in mathematics. Not so much in society. To go forward, go backward. What we want is a statement like, if a graph G is, what is good, then it is bipartite. So let's consider statements that begin, if a graph G is bipartite, then something. So I suppose G is a bipartite graph with disjoint vertex sets v1 v2. The vertices in any cycle must alternate between being in v1 and being in v2. Since a cycle has to begin where it ends, this means that if G is a bipartite graph, then every cycle in G has even length. In general, the converse of a true statement could be true or false, but in mathematical research it's a starting point. Suppose every cycle in G has even length. Do we know anything about G? Let's take a look at that next.